# Rindler coordinates in spherical form

From the Minkowski metric, it is almost straightforward to change coordinates and pass to Rindler metric $$ds^2 = e^{2a\xi}(-d\tau^2 + d\xi^2) + dy^2 + dz^2$$
given the transformation law $$t = \frac{1}{a}e^{a\xi}\sinh{a\tau}$$ $$x = \frac{1}{a}e^{a\xi}\cosh{a\tau}$$ At any fixed $$\xi$$, let's say $$\xi = 0$$, these coordinates represent the frame of an uniformly accelerated observer, with acceleration pointing in the x-direction. The question is how can i find a "polar" or rather "spherical" version of the Rindler coordinates describing the frame of an observer accelerating in the radial direction of a $$(t,r,\theta,\phi)$$ coordinate system? Writing Minkowski in spherical coordinates is very simple, but it is not trivial to make the same change for Rindler, due to this radial acceleration vs x-acceleration question.

• There is no such thing as the radial direction. Can you clarify what you mean? Personally I think you should be looking at cylindrical coordinates. Spherical coordinates don’t make sense as there isn’t any spherical symmetry even approximately
– Dale
Commented Jul 8, 2022 at 12:25

If I understand correctly, let $$(t,r,\theta,\phi)$$ be the spherical coordinates entered about an arbitrary origin of an inertial frame, you want a new coordinate system $$(\tau,\rho,\theta,\phi)$$ such that at fixed $$\rho_0,\theta_0,\phi_0$$, the worldline parametrised by $$\tau$$ is the one of an accelerating observer at proper rate $$a$$ in the radial direction with $$\tau$$ their proper time. For $$\rho$$, the natural parametrisation would be the point of closest approach, but to keep your approach, I'll take a normalised log of the distance of closest approach.
You just need to calculate position of these accelerating observers: $$t = \frac{1}{a}e^{a\rho_0}\sinh(a\tau) \\ r = \frac{1}{a}e^{a\rho_0}\cosh(a\tau) \\ \theta = \theta_0 \\ \phi = \phi_0 \\$$ which gives you your coordinate change. Plugging it into the metric: $$ds^2=-dt^2+dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi\\ =-e^{2a\rho}d\tau^2+e^{2a\rho}d\rho^2+\left(\frac{1}{a}e^{a\rho}\cosh(a\tau)\right)^2d\theta^2+\left(\frac{1}{a}e^{a\rho}\cosh(a\tau)\right)^2\sin^2\theta d\phi^2$$
Taking instead the distance of closest approach, which is physically more transparent and the analogy between Rindler coordinates and polar coordinates is made more apparent, your original formulas become: $$\tilde \xi = \frac{1}{a}e^{a\xi}\\ t = \tilde \xi\sinh(a\tau) \\ x = \tilde \xi\cosh(a\tau) \\ ds^2 = -(a\tilde \xi)^2d\tau^2+d\tilde\xi^2+dy^2+dz^2$$
so the new formulas become: $$\tilde \rho = \frac{1}{a}e^{a\rho}\\ ds^2=-(a\tilde\rho)^2d\tau^2+d\rho^2+\tilde \rho^2\cosh^2(a\tau)d\theta^2+\tilde\rho^2\cosh^2(a\tau)\sin^2\theta d\phi^2$$